Prove $ ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} )^{1/4} + ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} )^{1/4}\ge 68^{1/4}$ 
Let $0<\theta<\pi/2$. Prove that $$\left ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} \right )^{1/4}+\left ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} \right )^{1/4}\geqslant (68)^{1/4}$$
and find when the equality case holds.

This is a competition math problem. The material used should only cover up to pre-calculus.
So I quickly found out that equality holds when both of the $\sin^2(\theta)$ and $\cos^2(\theta)$ equals to 1/2, but I am not sure how to prove that this equality is true. I tried to substitute for variables and also use trig identities but just can't find out a way to do this. Thank you guys for helping me.
 A: By Minkowski (see here: https://en.wikipedia.org/wiki/Minkowski_inequality) we obtain
$$\sqrt[4]{ \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta}}+\sqrt[4]{ \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta}}\geq\sqrt[4]{\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4}.$$
Thus, it remains to prove that
$$\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4\geq68.$$
Now, let $\sin\theta+\cos\theta=2k\sqrt{\sin\theta\cos\theta}$.
Thus, $k\geq1$ and we need to prove that
$$\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2(\sin^2\theta+\cos^2\theta)^2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4\geq68(\sin^2\theta+\cos^2\theta)$$ or
$$(k+1)^2+16(2k^2-1)^2(k+1)^2\geq68(2k^2-1)$$ or
$$(k-1)(64k^5+192k^4+192k^3+64k^2-119k-85)\geq0,$$
which is obvious.
Done!
A: An alternative formulation:
Let $\frac{\sin^2 \theta}{2} = x$ and $\frac{\cos^2 \theta}{2} = y$
Minimize $f(x,y) = (x+\frac{1}{y})^{\frac{1}{4}}+(y+\frac{1}{x})^{\frac{1}{4}}$ subject to $x+y = \frac{1}{2}, x > 0, y > 0$
$f(x,y)$ has a local minimum at $x = y = \frac{1}{4}$
(easy to show using Lagrange multipliers)
Let $\mathcal{L} = (x+\frac{1}{y})^{\frac{1}{4}}+(y+\frac{1}{x})^{\frac{1}{4}} + \lambda (x+y- \frac{1}{2})$
Set $\frac{\partial \mathcal{L}}{\partial x} = \frac{\partial \mathcal{L}}{\partial y} = \frac{\partial \mathcal{L}}{\partial \lambda} = 0$ and solve.
The first two equations imply $x = y$ and from the third equation, we finally have $x = y = \frac{1}{4}$
A: Since you proved that the equality holds for $\sin^2(\theta)=\cos^2(\theta)=\frac 12$, in the same spirit as PTDS's answer, consider the function
$$f=\sqrt[4]{\frac{1-t}{2}+\frac{2}{t}}+\sqrt[4]{\frac{t}{2}+\frac{2}{1-t}}$$ and develop it as a Taylor series around $t=\frac 12$.
You should get a polynomial in $(t-\frac 12)^{2k}$ in which all coefficients are positive. Limited to very first orders
$$f= \sqrt[4]{68}+\frac{77 }{4 \sqrt{2}\,
   17^{3/4}}\left(t-\frac{1}{2}\right)^2+\frac{54803}{1088 \sqrt{2}\,
   17^{3/4}} \left(t-\frac{1}{2}\right)^4+O\left(\left(t-\frac{1}{2}\right)^6\right)$$ Then $\sqrt[4]{68}$ is the minimum value.
A: Clearly, we only need to prove the case when $\theta \in (0, \pi/4]$.
Let
\begin{align*}
 A &= \frac{4}{17}\left(\frac{\sin^2 \theta}{2} + \frac{2}{\cos^2\theta}\right), \\
 B &= \frac{4}{17}\left(\frac{\cos^2 \theta}{2} + \frac{2}{\sin^2\theta}\right).
\end{align*}
It suffices to prove that
$$A^{1/4} + B^{1/4} \ge 2$$
or
$$B^{1/4} - 1 \ge 1 - A^{1/4}$$
or
$$\frac{B - 1}{(B^{1/4} + 1)(B^{1/2} + 1)} \ge \frac{1 - A}{(1 + A^{1/4})(1 + A^{1/2})}$$
or
$$\frac{B - 1}{B^{3/4} + B^{1/2} + B^{1/4} + 1}
\ge \frac{1 - A}{1 + A^{1/4} + A^{1/2} + A^{3/4}}.$$
It is easy to prove that
$B \ge 1 \ge A$.
It suffices to prove that
$$\frac{B - 1}{B + B + B + 1}
\ge \frac{1 - A}{1 + A + A + A}$$
or
$$3AB - A - B - 1 \ge 0$$
or (clearing the denominators)
$$(3\sin^4\theta - 3\sin^2\theta + 56)(1 - 2\sin^2\theta)^2 \ge 0$$
which is true, with equality if $\sin^2\theta = 1/2$.
We are done.
